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Espectroscopia UV-vis e métodos quiroópticos

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O que você vai aprender

  • Assign UV-vis bands to π→π* and n→π* transitions from their position, intensity and solvent shifts, and choose a solvent and cuvette that are transparent where you measure.
  • Predict how conjugation, ring fusion, auxochromes and planarity move λmax, and estimate λmax of dienes and enones with stated Woodward–Fieser values.
  • Use the Beer–Lambert law to find concentrations, molar absorptivities and mixture compositions, and recognise when a measurement leaves the linear range.
  • Calculate specific rotation and enantiomeric excess, and interpret circular dichroism spectra of chiral chromophores and proteins.

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Electronic transitions and chromophores

Ultraviolet and visible light (≈200–800 nm) promotes an electron from a filled orbital to an empty one. The photon energy is E = hc/λ, so a shorter wavelength means a larger energy gap: 217 nm corresponds to about 551 kJ/mol. Saturated alkanes only have σ→σ* transitions below ≈150 nm, which is why hexane and cyclohexane are good UV solvents. A chromophore is the group that absorbs. A C=C or conjugated π system gives an intense π→π* band (ε ≈ 10³–10⁵ L mol⁻¹ cm⁻¹). A carbonyl group also has a weak n→π* band at longer wavelength: acetone absorbs at ≈279 nm with ε ≈ 15, because the oxygen lone pair lies in the molecular plane and overlaps poorly with the perpendicular π* orbital, so the transition is symmetry-forbidden. Polar, hydrogen-bonding solvents lower the energy of the lone pair and push n→π* bands to shorter wavelength (acetone ≈265 nm in water), whereas π→π* bands usually move slightly to longer wavelength.

Shift vocabulary, auxochromes and practical limits

A shift to longer wavelength is bathochromic (red shift), to shorter wavelength hypsochromic (blue shift); a rise in ε is hyperchromic, a fall hypochromic. An auxochrome is a substituent with lone pairs that conjugates with a chromophore. Benzene's weak band at ≈255 nm (ε ≈ 200) moves to ≈270 nm in phenol (ε ≈ 1450) and to ≈280 nm in aniline (ε ≈ 1430). Protonating aniline ties up the nitrogen lone pair, and anilinium absorbs at ≈254 nm, like benzene; deprotonating phenol frees a second lone pair and phenoxide absorbs at ≈287 nm. Measure only where the solvent and the cell are transparent: acetone cuts off near 330 nm, benzene near 280 nm and chloroform near 245 nm, and glass or plastic cuvettes absorb below about 320 nm, so work below that needs quartz (fused silica). A coloured compound absorbs in the visible and shows the complementary colour.

Conjugation moves λmax to longer wavelength

Each extra conjugated C=C raises the HOMO and lowers the LUMO, so the gap shrinks and λmax grows: ethene ≈165 nm, buta-1,3-diene 217 nm, hexa-1,3,5-triene ≈258 nm, and β-carotene, with eleven conjugated C=C, ≈450 nm (orange). Fused aromatics follow the same trend: the longest-wavelength band moves from ≈255 nm (benzene) to ≈312 nm (naphthalene), ≈375 nm (anthracene) and ≈471 nm (tetracene, orange). Non-conjugated C=C groups absorb as separate alkenes, below 200 nm. Conjugation needs overlap: trans-stilbene is planar and absorbs at ≈295 nm (ε ≈ 27 000), while cis-stilbene is twisted by steric clash and absorbs at ≈280 nm (ε ≈ 10 500). Ortho substituents that twist biphenyl out of plane remove its ≈250 nm conjugation band.

Woodward–Fieser estimates

Empirical rules predict λmax of the π→π* band of dienes and enones within a few nanometres. The values used in this topic: dienes, parent 214 nm for an acyclic or heteroannular diene (the two C=C in different rings) and 253 nm for a homoannular diene (both C=C in one ring); add 5 nm for each alkyl group or ring residue on a diene carbon, 5 nm for each exocyclic C=C and 30 nm for each additional conjugated C=C. Enones, parent 215 nm for an acyclic or six-membered-ring enone; add 10 nm for an α alkyl group or ring residue, 12 nm for β, 18 nm for γ or δ, 5 nm for each exocyclic C=C and 30 nm for each additional conjugated C=C. A C=C is exocyclic when one of its carbons belongs to a ring that the other carbon is not in. In the octalone drawn here the enone parent is 215, the β carbon carries two ring residues (+24) and the C=C is exocyclic to the second ring (+5): 244 nm.

The Beer–Lambert law

Absorbance A = log₁₀(I₀/I) = εlc, where ε is the molar absorptivity (L mol⁻¹ cm⁻¹), l the path length (cm) and c the concentration (mol/L). Transmittance T = I/I₀ = 10⁻ᴬ, so A = 1 means 10% of the light gets through and A = 2 only 1%. Absorbances of independent absorbers add, which lets two wavelengths resolve a two-component mixture. Measure at λmax, where sensitivity is highest and the band is flat, and keep A roughly between 0.1 and 1: at high absorbance so little light reaches the detector that stray light and noise bend the calibration line. When spectra recorded during a clean A ⇌ B conversion all cross at one wavelength, that isosbestic point marks equal ε for the two species.

Optical rotation and enantiomeric excess

A chiral, non-racemic sample rotates plane-polarised light. Specific rotation [α] = α/(l·c), with α the observed rotation in degrees, l in decimetres and c in g/mL, usually at the sodium D line (589 nm) and a stated temperature and solvent. Enantiomers have equal and opposite [α]; racemates and meso compounds give zero. The sign is an experimental quantity and cannot be read from the R/S descriptor: (R)-glyceraldehyde is dextrorotatory, (R)-butan-2-ol levorotatory. Enantiomeric excess ee = (major − minor)/(major + minor); a 90:10 mixture has 80% ee. If rotation is linear in composition, optical purity [α]obs/[α]pure equals ee. Rotation-based ee is easily distorted by achiral impurities or solvent left in the sample, so chiral HPLC or GC is preferred. A polarimeter cannot distinguish rotations that differ by a multiple of 180°; measuring again at a suitably chosen lower concentration resolves the ambiguity.

Circular dichroism and optical rotatory dispersion

Circular dichroism (CD) measures Δε = εL − εR, the difference in absorption of left- and right-circularly polarised light. A CD band appears only where a chiral sample absorbs, so a chiral ketone shows its n→π* Cotton effect near 290 nm while a chiral alkane shows nothing above 200 nm. Optical rotatory dispersion (ORD) is the variation of rotation with wavelength; it is non-zero even far from absorption bands, which is why a chiral alkane still has a measurable [α]D. Enantiomers give mirror-image CD spectra of equal magnitude, and racemic or achiral samples give none. In proteins the amide chromophores report secondary structure: an α-helix shows negative bands at 222 and 208 nm and a positive band near 192 nm, a β-sheet a single negative band near 217 nm, and a disordered chain a strong negative band near 198 nm.

What to carry forward

Find the chromophore, then reason about its gap: more conjugation, planar overlap and lone-pair donors push λmax to longer wavelength; n→π* bands are weak and move to shorter wavelength in protic solvents. Use stated Woodward–Fieser values to estimate λmax, A = εlc for quantitation inside the linear range, [α] = α/(l·c) and ee = (major − minor)/total for chiral samples, and remember that CD needs a chiral chromophore absorbing at the wavelength you measure.

Problemas de prática (40)

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Preveja o produto 3

Optical rotation, ee and CD

  • The CD spectrum of the other enantiomerPredict the CD spectrum of an enantiomer from that of its mirror image.

Conjugation and λmax trends

  • Which pure hydrocarbon is coloured?Predict which aromatic hydrocarbon absorbs visible light from the extent of its conjugated π system.

Chromophores and electronic transitions

  • Aniline's band in acidPredict how protonating an amine auxochrome changes an aromatic absorption band.

Classificar 3

Optical rotation, ee and CD

  • Which sample gives a CD band near 290 nm?Classify samples by whether they meet both requirements for a CD band: chirality and absorption at the wavelength measured.
  • Which sample rotates plane-polarised light?Classify pure compounds as optically active or inactive by recognising chirality, meso compounds and achiral molecules.

Chromophores and electronic transitions

  • Name the shift from benzene to phenolClassify a change in band position and intensity with the bathochromic/hypsochromic and hyperchromic/hypochromic terms.

Comparar 5

Beer–Lambert quantitation

  • Rank solutions by absorbanceRank solutions by absorbance by combining molar absorptivity, path length and concentration.

Conjugation and λmax trends

  • Rank linear acenes by longest absorptionOrder linearly fused aromatic hydrocarbons by the wavelength of their longest absorption band.
  • Which C₆H₁₀ isomer absorbs above 200 nm?Identify the conjugated isomer among C₆H₁₀ dienes and cycloalkenes from its UV absorption.
  • Rank polyenes by λmaxOrder polyenes by λmax from the number of conjugated C=C bonds.
  • Longest λmax and largest ε among stilbene relativesCompare how planarity and conjugation between two aryl rings affect both λmax and ε.

Calcular 15

Beer–Lambert quantitation

  • Concentration from absorbanceRearrange A = εlc to find a concentration from a measured absorbance.
  • Molar absorptivity from a weighed sampleDetermine a molar absorptivity from a weighed mass, a volumetric dilution and a measured absorbance.
  • NADH concentration after a dilutionCorrect a Beer–Lambert concentration for a volumetric dilution made before the measurement.
  • Percent transmittance from absorbanceConvert an absorbance into percent transmittance with T = 10⁻ᴬ.
  • Analyse a two-component mixtureResolve the concentration of one component of a mixture from absorbances at two wavelengths using additivity.

Optical rotation, ee and CD

  • Enantiomer percentage from rotationConvert an observed specific rotation into the percentage of the major enantiomer.
  • ee after mixing with racemateCalculate the enantiomeric excess of a blend of enantiopure material and racemate.
  • Specific rotation from a polarimeter readingCalculate a specific rotation from an observed rotation, a path length in decimetres and a concentration in g/mL.

Chromophores and electronic transitions

  • Energy of the butadiene π→π* photonConvert an absorption wavelength into a molar transition energy with E = N_A·hc/λ.

Woodward–Fieser predictions

  • Woodward–Fieser estimate for a dienoneExtend the enone rules to a linear dienone by adding the extra-conjugation and δ-substituent increments.
  • Woodward–Fieser estimate for 2,3-dimethylbutadieneApply the diene parent value and alkyl increments to an acyclic conjugated diene.
  • Woodward–Fieser estimate for a heteroannular dieneCount ring residues and exocyclic double bonds to estimate λmax of a heteroannular diene in a fused bicyclic ring system.
  • Woodward–Fieser estimate for a homoannular dieneEstimate λmax of a homoannular diene in a fused bicyclic ring system from the parent value, ring residues and an exocyclic double bond.
  • Woodward–Fieser estimate for mesityl oxideApply the enone parent value and β-substituent increments to an acyclic α,β-unsaturated ketone.
  • Woodward–Fieser estimate for a bicyclic enoneEstimate λmax of a fused bicyclic enone, recognising β ring residues and an exocyclic C=C.

Encontre o erro 3

Beer–Lambert quantitation

  • An unknown above the calibrated rangeDiagnose why an absorbance far above the calibrated range gives an unreliable concentration and choose the remedy.

Optical rotation, ee and CD

  • Rotation and HPLC disagree on eeDiagnose why a rotation-based optical purity falls below the ee measured by chiral HPLC.

Chromophores and electronic transitions

  • A blank that absorbs at 260 nmDiagnose an off-scale UV blank from the transmission range of the cuvette material.

Interprete as evidências 3

Beer–Lambert quantitation

  • Spectra that cross at one wavelengthInterpret an isosbestic point in a series of spectra recorded during an acid–base titration.

Optical rotation, ee and CD

  • Read a far-UV protein CD spectrumInterpret a far-UV CD spectrum of a protein in terms of its dominant secondary structure.

Conjugation and λmax trends

  • Assign the two bands of cyclohexenoneAssign the π→π* and n→π* bands of an α,β-unsaturated ketone from position and intensity.

Trabalhe de trás para frente 1

  • Which hexahydronaphthalene absorbs at 275 nm?Work backwards from an observed λmax to the isomer whose Woodward–Fieser estimate matches it.

Planeje uma síntese 2

Optical rotation, ee and CD

  • Resolve an ambiguous polarimeter readingPlan a polarimetry measurement that removes the 180° ambiguity in an observed rotation.

Chromophores and electronic transitions

  • Choose a solvent for a 220 nm measurementChoose a UV solvent by recognising which candidates contain chromophores that absorb at the measuring wavelength.

Explique por quê 5

Optical rotation, ee and CD

  • Rotation without a CD bandExplain why a chiral compound can have a non-zero [α]D but no CD band in the near-UV and visible.
  • Does R mean dextrorotatory?Explain why the sign of optical rotation cannot be predicted from the R/S descriptor.

Conjugation and λmax trends

  • Why 2,2′-dimethylbiphenyl loses its strong bandExplain how steric twisting about a single bond removes a conjugation band.

Chromophores and electronic transitions

  • Why acetone's weak band moves in waterExplain the blue shift of a carbonyl n→π* band in a hydrogen-bonding solvent.
  • Why the carbonyl n→π* band is weakExplain the low intensity of the carbonyl n→π* band from the orientation of the orbitals involved.

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